$\int x^3e^{x^4}dx\,= $ $+~C$
Notice that we can rewrite the integral as $ \int e^{x^4}\cdot \,x^3\, dx\,$. If we let $ {u=x^4}$, then ${du=4x^3 \, dx}$ and $x^3\, dx=\dfrac{du}{4}}$. Substituting gives us: $\begin{aligned} \int e^{{x^4}}\cdot \,x^3\, dx}\,&= \int e^{ u}\, \cdot \dfrac{du}{4}}\,\\\\\\ &= \int e^u\cdot \dfrac14\, du\\\\\\ &=\dfrac14 \int e^u\, du\end{aligned}$ We recognize this antiderivative. $\begin{aligned}\phantom{\int\dfrac{e^x}{1+e^{2x}}dx}&=\dfrac14 \int e^u\, du~\\\\\\ &=\dfrac14e^u+C\end{aligned}$ We can now substitute back to find the antiderivative in terms of $x$. ∫ e x 1 + e 2 x d x = 1 4 e u + C = 1 4 e x 4 + C \begin{aligned}\phantom{\int\dfrac{e^x}{1+e^{2x}}dx~}&=\dfrac14e^u+C\\\\\\\ &=\dfrac14e^{ x^4}+C\end{aligned} The answer: $\int x^3e^{x^4}dx\,= \dfrac14e^{ x^4}+C$